What is the solution for y^{''}-2y^'+5y=0,y(pi/2)=0,y^'(pi/2)=2 ?

The solution for y^{''}-2y^'+5y=0,y(pi/2)=0,y^'(pi/2)=2 is y=-e^{(2t-pi)/2}sin(2t)

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y^{''}-2y^'+5y=0,y(pi/2)=0,y^'(pi/2)=2

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 2

y = - e^{\frac{2 t - π}{2}} sin (2 t)

Solution steps

y^{''} (t) - 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = - e^{\frac{2 t - π}{2}} sin (2 t)

y = - e^{\frac{2 t - π}{2}} sin (2 t)

x \to 0 lim ((ln (x)) - \frac{1}{x})

x \to - 2 lim (- (x^{2} - x + 1))

\int_{1}^{2} x^{10} d x

y = \frac{5}{x ^{2}}

ln (e^{7 x - 2})

What is the solution for y^{''}-2y^'+5y=0,y(pi/2)=0,y^'(pi/2)=2 ?
The solution for y^{''}-2y^'+5y=0,y(pi/2)=0,y^'(pi/2)=2 is y=-e^{(2t-pi)/2}sin(2t)